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A big THANK YOU for the very good site. HERE IS A SOLUTION for Euklids book XIII 10. A very clever inviting proof. I think the german words are easy to understand. But its not a "classical" answer - have you one?

See attached picture in the hyperlink belowhttp://yfrog.com/6wproblem533jLet H is the intersection of OA to BD and E is the symmetric of A over BDLet x=BE=BA=DA and y=BH=HD=1/2BD=1/2ACNote that angle DBA=angle EBH=18 , angle EOB=angle OBE= 36 and BE is the angle bisector of angle OBA

Since BE is the angle bisector so we have x/R=(R-x)/x or R.x=R^2-x^2In right triangle OHB we have BH^2=OB^2-OH^2 or y^2=R^2-(R+x)^2/4 or 4.y^2=3.R^2-2.R.x-x^2Replace R.x=R^2-x^2 in above expression we have 4.y^2=R^2 +x^2 And AC^2=BC^2+AB^2 . Triangle ABC is a right triangle

TO prove the proposition we need to see that in the given figure if we apply Pythagoras theorem then we shall get〖 4R〗^2 〖sin〗^2 18+R^(2 )=4R^2 〖sin〗^2 36.Then we need to see whether the values of sin 18 and sin 36 satisfy the above relation.For this we need to simplify the expression after which we shall get 〖sin〗^2 36-〖sin〗^2 18=1/4. to see really if it is true then we substitute θ in place of 18 and 36 respectively and then solve the equation〖 sin〗^2 2θ-〖sin〗^2 θ=1/4. By using identity we get sin3θ.sinθ= 1/4. which then simplifies to 3〖sin〗^2 θ-4〖sin〗^4 θ= 1/4 proceeding step by step we get the following:12〖sin〗^2 θ-16〖sin〗^4 θ=1. ⇒ 12〖sin〗^2 θ=1+16〖sin〗^4 θ ⇒ 12〖sin〗^2 θ=〖(1+〖4sin〗^2 θ)〗^2- 8〖sin〗^2 θ ⇒20〖sin〗^2 θ=〖(1+〖4sin〗^2 θ)〗^(2 ) ⇒2√5 sinθ= 1+4〖sin〗^2 θ ⇒4〖sin〗^2 θ-2√5 sinθ+1=0. solving this equation as quadratic in sin θ we get sin θ = (√5 ±1)/4 now we get the value of θ as 18 degrees (by applying inverse function of sin) and that of 3θ as 54 degrees. So it shows that the expression 〖 4R〗^2 〖sin〗^2 θ+R^(2 )=4R^2 〖sin〗^2 θ is satisfied by the obtained values of θ. Thus the proposition is proved.Q. E. D.